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scoray [572]
4 years ago
10

Blow up a balloon and rub it against your shirt a number of times. In doing so you give the balloon a net electric charge. Now t

ouch the balloon to the ceiling. On being released, the balloon will remain stuck to the ceiling. Why?
Physics
1 answer:
tamaranim1 [39]4 years ago
8 0
<span>Balloons are blown up, and then rubbed against your shirt many times. The balloon then touches the ceiling. When released, the balloon remains stuck to the ceiling. The balloon is charged by contact. The ceiling has a neutral charge. The charged balloon induces a slight surface charge on the ceiling opposite to the charge on the balloon. Balloon and ceiling electric charges are opposite in sign, so they will attract each other. Since both the balloon and the ceiling are insulators, charge can not flow from one to the other. The charge on the balloon is fixed on the balloon and the charge on the ceiling remains fixed to the ceiling. It just so happens that the<span> electrostatic force the ceiling exerts on the balloon is sufficient to hold the balloon in place (i.e. overcomes gravity, etc.).</span></span>
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Answer:

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R = X_{L} = j\omega L = 2\pi fL

where

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Now,

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C ∝ \frac{1}{x}

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\frac{Z}{I} = \frac{Z'}{I'}

\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}

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{R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})

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